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find all positive integer values of
inSorry there was an typo error . Now i have edited.
Last edited by juantheron (2011-10-19 08:26:56)
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I don't know if you're allowed to do this but this problem can be brute-forced pretty easily. The denominators will very quickly overcome the numerators and the left side will strictly decrease, so you don't have to check very many values.
Wrap it in bacon
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Hi juantheron;
For what it is worth, Mathematica quickly says no solutions. Welcome to the forum.
A graph of the function suggests this.
http://www.mathsisfun.com/graph/functio … ymax=3.753
2 and 4 are close but...
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Now I have edit my question.
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Hi;
You will find x = 4 now.
To prove there are no more roots we will use the notion of strictly decreasing as the TheDude suggests.
Theorem A) If f'(x) is strictly negative then f(x) is strictly decreasing.
That is proved here:
http://www.dpmms.cam.ac.uk/~wtg10/meanvalue.html
Differentiate each term:
This is strictly negative when x > 5.
This is strictly negative when x > 8
This is strictly negative when x > 1
So by the theorem A all three terms are strictly decreasing and f(x) is strictly decreasing. You now only need to try integer values for x = 1,2,3,4,5,6,7,8. You will find x= 4 and no others.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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thanks bobbym. got it.
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Hi juantheron;
Your welcome and welcome to the forum.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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